1 - Fluid - Continuity and Conservation of Matter - 2020 - 21

Introduction
CE 256: Fluid Mechanics

➢ Lectures
➢ Laboratory work
Assessment:
➢ 70% Examination
➢ 30% Continuous assessment
Continuous Assessment:
➢ Mid-Sem Examination
➢ Lab work (To be defended)
➢ Attendance in class
➢ Assignments
➢ Many others
Assignment
Make a comprehensive note on the following in two copies:
A copy for yourself as a note towards the end of
SEMESTER EXAMINATION.
A copy to be submitted as an ASSIGNMENT not later
than 12:00 GMT on Monday, 7th of June, 2021.
a. Uniform flow; Non-uniform; Steady flow; Unsteady
flow
b. Steady uniform flow; Steady non-uniform flow;
Unsteady uniform flow; Unsteady non-uniform flow.
c. Compressible fluid; Incompressible fluid
d. One dimensional flow; Two-dimensional flow
e. Streamlines; Streamtubes
FLUID DYNAMICS
Fluid Dynamics is a sub-discipline of Fluid

Mechanics that deals with fluid flow.
Sub-disciplines of fluid dynamics:

➢ Aerodynamics (Study of air and other gasses
in motion.
➢ Hydrodynamics (Study of liquids in motion)
Conclusion: Study and analysis of fluid in

motion.
Continuity and Conservation of Matter
• Basic Terms to Note
M = Mass flow rate, [M/T]
Q = Volume flow rate, [L / T]
3
W = Weight flow rate, N/s
➢ Mass flow rate

Mass of fluid
Mass flow rate ( M ) =
Time taken to collect that mass of fluid
Example: An empty tank weighs 10.0kg. After 20 minutes

of collecting water, the tank weighed 100.0kg.
Determine the Mass flow rate.
100.0 − 10.0
Mass flow rate = M = = 7.5 10-2 kg / s
20  60
Question
The Mass flow rate of a fluid from a pipeline is
8.5 kg/s. Determine the time it will take to fill a
container with 105 kg of the fluid.
mass
Time =
mass flow rate
105
=  12.4 s
8.5
➢ Volume flow rate (Discharge)
Volume of fluid flowing per unit time.
Volume of fluid
Discharge= Q =
Time taken tocollect that volumeof fluid
➢ Weight flow rate

Weight of fluid flowing per unit time.
Weightof fluid
W=
Time taken tocollect that weight of fluid
Note the following relations.
M = Q
W = Q
Example
If it takes 10 min to fill a reservoir of dimension
500 cm by 1000 cm by 200 cm from a pipe. What
is the discharge of the pipe in m3/s.
Volume 5 10  2
Discharge = =
Time 10  60
100
= = 0.17 m / s
3
600
If the density of a fluid is 1000 kg/m3 and the
mass flow rate ( M ) = 0.72kg / s.
Determine the Discharge?
Mass flow rate ( M ) = Q

Hence
M
0.72
Q= = = 7.0  10 m / s
-4 3
 1000
➢ Discharge and mean velocity
▪ Flow through a pipe
x
u
Pipe um umax
• Velocity in the pipe is not constant across the pipe cross
section.
• Velocity is zero at the walls and increases to a maximum at
the centre, then decreases symmetrically to the other
wall.
• Variation of velocity across the section is known as the
velocity profile or distribution.
Area, A x umt
x Area, A
Pipe Cylinder of fluid
Let
• Cross sectional area of a pipe at point X be A
• Mean velocity at section X to be um

• During time t, a cylinder of fluid will pass section x-x.
• Volume of fluid (cylinder) is given as Aumt.

Discharge (Q) = Volume per unit time
Volume A x um x t
Q= = The idea, of mean velocity
Time t multiplied by the area gives
the discharge, applies to all
Q = Aum or Q = Au situations - not just pipe flow
If
• Cross-sectional area, A of a pipe is 1.2 x 10-3m2
• Discharge, Q is 4.8 l/s,
Then the Mean velocity, um, of flow is

Q 4.8x10-3
um = = = 4.0 m/s
A 1.2x10-3
➢ Continuity
• Principle of conservation of mass: Matter cannot

be created or destroyed - (it is simply changed
into a different form of matter).
Applying this principle to fixed volumes (control
volumes, surfaces, fixed region)
• In General, for any control volume,
Mass entering Mass leaving  Change of mass in the control

= +
per unit time  per unit time  volume per unit time 

• For steady flow, no change in mass in the control

volume,
Changeof mass in the control

 =0
volume per unit time 
Applying the principle of conservation of mass to
analyse flowing fluids.
• Consider a streamtube (a pipe).

❖ No fluid flows across the boundary made by the
streamlines.
❖ Mass enters the streamtube at point 1 and only
leaves through point 2 of the streamtube.
2 A2 u2
1
A1
u1
This means for steady state,
mass enteringper  Mass leaving per 
 = 
unit time at end 1  unit time at end 2 
1Q1 = 2Q2
1 A1u1 =  2 A2 u2
• For the existence of steady flow,
1 A1u1 =  2 A2 u2 = constant
• For steady flow, quantity of fluid passing any section
(point) will remain constant (the same).
Ø = 100mm Ø = 50mm
1 2
• No fluid is added/removed between points 1 & 2

M1 = M 2 or 1A1u1 = 2 A2u2
For incompressible fluid (density constant), 1 =  2 = const.
A1u1 = A2u2 = Q1 = Q2 = constant Continuity equation

Example: Fluid with the following parameters.
T = 70o C, u1 = 8 m/s, Ø= Ø=
 70 = 9.59kN/m ; 70 = 978kg/m
3 3 100mm 50mm
1 2
Find u2, Q2 , W2 , M 2
Soln.
Using Continuity Equation
A1
A1u1 = A2u2 ; u2 = ( )u1
A2
D 2
 100 2
 502
A1 = 1
= = 7854 mm 2 A2 = = 1963 mm 2
4 4 4
Hence 7854
u2 =  8m/s = 32 m/s
1963
Discharge

2  1m 2 
Q1 = Q2 = A1u1 = A2u2 = (1963m ) (32m/s) = 0.063m 3 /s
( )
 103 mm 2 
 
Weight flow rate
W = Q;  70 = 9.59kN/m 3 ;
 W = (9.59kN/m )(0.063m /s) = 0.60kN/s
3 3
Mass flow rate

M = Q;  70 = 978kg/m3 ;
 M = (978kg/m3 )(0.063m 3 /s) = 616kg/s
Some applications of the continuity equation
❖A pipe which expands or diverges
❖Determination of velocities in pipes coming from a

junction.
Q1 = Q2 + Q3
A1u1 = A2 u2 + A3u3
Exercise
• If pipe 1 diameter = 150mm, mean velocity
4.2m/s, pipe 2 diameter 100mm takes 30% of
total discharge and pipe 3 diameter 60mm.
What are the values of discharge and mean
velocity in each pipe?

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Introduction

CE 256: Fluid Mechanics

Fluid Dynamics is a sub-discipline of Fluid

Sub-disciplines of fluid dynamics:

Conclusion: Study and analysis of fluid in

W = Weight flow rate, N/s

➢ Mass flow rate

Example: An empty tank weighs 10.0kg. After 20 minutes

Volume of fluid flowing per unit time.

➢ Weight flow rate

Determine the Discharge?

Mass flow rate ( M ) = Q

• Mean velocity at section X to be um

• Volume of fluid (cylinder) is given as Aumt.

Then the Mean velocity, um, of flow is

• Principle of conservation of mass: Matter cannot

Mass entering Mass leaving  Change of mass in the control

• For steady flow, no change in mass in the control

Changeof mass in the control

• Consider a streamtube (a pipe).

• For the existence of steady flow,

• No fluid is added/removed between points 1 & 2

A1u1 = A2u2 = Q1 = Q2 = constant Continuity equation

Weight flow rate

Mass flow rate

❖A pipe which expands or diverges

❖Determination of velocities in pipes coming from a

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